# The formula for the sum of an arithmetic series

The purpose of this lesson is to find the sum of an arithmetic series. An arithmetic series is a series of numbers t1, t2, t3, ...tn, such that the difference "d" between any two adjacent numbers, like t3 and t4, or t8 and t9, is always the same.

In other words, if there is a number d, such that the series can be written as t1, t1 +d, t1+2d, t1+3d, ..., then the series is an arithmetic series.

The simplest and best-known example is to simply count from 1 to 10: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. You can easily see that the difference between any two adjacent terms is 1.

To add them up, we don't need a formula. But what if the number of terms is much higher, perhaps as high as 100? Then it is a lot of work and easy to make a mistake. Here a formula would help.

# becomes famous mathematician.

This is exactly the situation faced by Gauss, the famous mathematician, when he was a kid. His teacher told the class to add up all the numbers from 1 to 100 and report the result to him. The only student who got it right was Gauss, and he did not have to do much calculating, because he found the formula we are looking for.

To illustrate the basic idea, take a look at this example: S = 1 + 2 + 3 + 4 + 5 + 6 Again, we don't need a formula. It is easy to add them up and get 21. But there is an easier way. This is how it works.

You should already know that adding numbers can be done in any order, so the sum of 3 + 9 + 5 is the same as the sum of 5 + 3 + 9. Because of this, we can write the terms backwards and still get the same result: S = 6 + 5 + 4 + 3 + 2 + 1

Now add them up vertically, term by term like this:

S = 1 + 2 + 3 + 4 + 5 + 6
S = 6 + 5 + 4 + 3 + 2 + 1
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S + S = 7 + 7 + 7 + 7 + 7 + 7

By pairing the terms like this we get a series of 6 sums of pairs, where the sum of each pair is always the same: 7. We can now calculate the sum on the right as 7 times 6, which is equal to 42. On the left side of the equation, we have S + S, or 2S. Since 2S = 42, we can easily see that S must be equal to 21.

Now let's try the same idea with 10 terms:

S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10
S = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
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S + S = 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11 + 11

Here we get 10 pairs of 11, so 2S = 110, giving S = 55. Let's try it with 100 terms, but I am only writing out the first and last three terms

S = 1 + 2 + 3 + ... + 98 + 99 + 100
S = 100 + 99 + 98 + ... + 3 + 2 + 1
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S + S = 101 + 101 + 101 + ... + 101 + 101 + 101

Since the number of terms is 100 and each pair sums to 101, we get 2S = 100 * 101, or S = 10100/2=5050

Now we know that this is a great idea, but it would be even greater if we had a formula, where you just plug in the number of terms. Let's see if we can get there: Let's call the number of terms n. We have seen that when n is 6, the sum is equal to: (7 * 6)/2, or ((n+1)*n))/2 when n is 10, the sum is equal to: ( 11 * 10)/2, or ((n+1)*n))/2 when n is 100, the sum is equal to: ( 101 * 100)/2, or ((n+1)*n))/2 So now we have a formula that works for any number of terms.

# Generalizing Gauss's formula

But what about a series like this: 4, 7, 10, 13, 16? This is also an arithmetic series. How do we know? We know because the difference between any two adjacent terms is always 3. Adding them up is not hard, we get 50.

Once again, we can write the sum two ways, like Gauss did, forwards and backwards, and add the terms in pairs:

S = 4 + 7 + 10 + 13 + 16
S = 16 + 13 + 10 + 7 + 4
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S + S = 20 + 20 + 20 + 20 + 20

We get 2S = 20 * 5 = 100, or S = 50.

For this kind of arithmetic series, we need a different formula. Remember the definition of an arithmetic series: a series of numbers where the difference between any two adjacent numbers is always the same. The series does not have to start with 1. Ours starts with 4. The difference between adjacent numbers does not have to be 1. In our latest example it is 3. If we call the first term "a", and the difference between adjacent numbers "d", then any arithmetic series t1, t2, t3, t4, ... can be written as a, a+d, a+2d, a +3d, ...

Note that the last term in a series of n terms is a + (n-1)*d. You can check this easily by plugging in some numbers. If a=1, d=1 and n=5, we get the series 1, 2, 3, 4, 5. The last term is 5, which is the same as a + (n-1)*d, since a=1, d=1 and n=5 If a=4, d=3 and n=5, we get the series 4, 7, 10, 13, 16. The last term is 16, which is the same as a + (n-1)*d, since a=4, d=3 and n=5

Then we can again rewrite it as Gauss did:

S = a + (a+d) + (a+2d) + ... + (a + (n-3)*d) + (a + (n-2)*d) + (a + (n-1)*d)
S = (a + (n-1)*d) + (a + (n-2)*d) + (a + (n-3)*d) + ... + (a +2d) + (a + d) + a
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S + S = (2a + (n-1)*d) + (2a + (n-1)*d) + (2a + (n-1)*d) + ... + (2a + (n-1)*d) + (2a + (n-1)*d) + (2a + (n-1)*d)

We get 2S = (2a + (n-1)*d)*n or S = (2a + (n-1)*d)*n/2, which is our new formula.

We can test it with our series, where a=4, d=3 and n=5. We get S=(2*4 + (5-1)*3)*5/2 = (8 + 4*3)*5/2 = 20 * 5 / 2 = 50 We can simplify this a bit, by noting that the last term, tn, is a + (n-1)*d Since S = (2a + (n-1)*d)*n/2 = (a + a + (n-1)*d)*n/2 = (a + tn)*n/2

In its most general and simplest form, the sum of an arithmetic series can be calculated like this: 1.) take the average of the first and last term, i.e (a + tn)/2, and 2.) multiply it by n, the number of terms. This is as simple as it gets: For the numbers 1 to 100, step 1, take the average of the 1st and last term and you get (1+100)/2=50.5 step 2, multiply the number from step 1 by the number of terms, which is 100, and you get 5050. For the series 4, 7, 10, 13, 16, do the same thing: The average of 4 and 16 is 10. The number of terms is 5. Multiply those two numbers and you get 50.

# Reviewing what we've learned

To review this lesson, you should now know that: 1.) an arithmetic series is a series of n numbers, t1, t2, t3, ..., tn, where any pair of adjacent terms differ by the same amount d. 2.) To find the sum of all its terms, you just take the average of its first term, t1, and its last term, tn, and multiply this average by n, the number of terms in the series.

To do well on the quiz for this lesson, you also need to know this: 1.) For the formula, we don't need to know the value of d, only the values of t1, tn, and n. 2.) We can easily calculate the value of d, if we know the values of all the other three. 3.) If we know only two of the three parameters (t1, tn, and n) required by the formula, but we also know d, then we can calculate the missing parameter (t1, tn or n). 4.) The difference, d, between any two adjacent terms does not have to be 1. 5.) The first term, t1, does not have to be 1.